feat: add 050

Author: Blankj

README.md

@@ -66,6 +66,7 @@
 |33|[Search in Rotated Sorted Array][033]|Arrays, Binary Search|
 |43|[Multiply Strings][043]|Math, String|
 |49|[Group Anagrams][049]|Hash Table, String|
+|50|[Pow(x, n)][050]|Math, Binary Search|
 |554|[Brick Wall][554]|Hash Table|
 
 
@@ -129,6 +130,7 @@
 [033]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/033/README.md
 [043]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/043/README.md
 [049]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/049/README.md
+[050]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/050/README.md
 [554]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/554/README.md
 
 [004]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/004/README.md

note/050/README.md

@@ -0,0 +1,39 @@
+# [Pow(x, n)][title]
+
+## Description
+
+Implement pow(x, n).
+
+**Tags:** Math, Binary Search
+
+
+## 思路0
+
+题意是让你计算`x^n`，如果直接计算肯定会超时，那么我们可以想到可以使用二分法来降低时间复杂度。
+
+```java
+class Solution {
+    public double myPow(double x, int n) {
+        if (n < 0) return helper(1 / x, -n);
+        return helper(x, n);
+    }
+
+    private double helper(double x, int n) {
+        if (n == 0) return 1;
+        if (n == 1) return x;
+        double d = helper(x, n >>> 1);
+        if (n % 2 == 0) return d * d;
+        return d * d * x;
+    }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/powx-n
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

src/com/blankj/medium/_050/Solution.java

@@ -0,0 +1,29 @@
+package com.blankj.medium._050;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/10/18
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+    public double myPow(double x, int n) {
+        if (n < 0) return helper(1 / x, -n);
+        return helper(x, n);
+    }
+
+    private double helper(double x, int n) {
+        if (n == 0) return 1;
+        if (n == 1) return x;
+        double d = helper(x, n >>> 1);
+        if (n % 2 == 0) return d * d;
+        return d * d * x;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.myPow(8.88023, 3));
+    }
+}